Summary and Analysis

Study Tools

Optical Phenomena

Interference

Diffraction

Problem :
What is the amplitude of the magnetic field of a light ray that has an irradiance of 0.00003 W/cm
2
.

We know that
I =
. If we assume the magnetic field is harmonic then
< B2 > = < B02cos2(σt) >
, but the time integral of the cosine squared is just
1/2
so this is equal to
B02/2
. Thus substituting and rearranging the first equation,
B0 = = = 5.0×10-8
tesla.

Problem :
What is the position of the fourth maximum for a double-slit apparatus with slits 0.05 centimeters apart and
a screen 1.5 meters distant when performed with monochromatic red light of frequency
384×1012
Hz?

The wavelength of this light is
λ = c/ν = 7.81×10-7
meters. Just plugging to the
formula
ym = = = 9.38
millimeters from the central bright maximum.

Problem :
In a Young's Double Slit experiment, what is the ratio of the irradiance at a distance 1 centimeter from
the center of the pattern, irradiance of each individual beam entering through the slits (assume the same
set up as before: light of frequency
384×1012
Hz, 0.05 centimeters between the slits, and a
screen 1.5 meters away)?

The irradiance as a function of distance from the center of the pattern is given by
I = 4I0cos2
, where
I0
is the irradiance of each of the interfering
rays. Plugging into the formula:
I = 4I0cos2() = 1.77I0
. Thus the ratio is just 1.77.

Problem :
A stream of electrons, each having an energy 0.5 eV is incident on two extremely thin slits 10
-2
millimeters apart. What is the distance between adjacent minima on a screen 25 meters behind the slits (
me = 9.11×10-31
kilograms, and
1eV = 1.6×10-19
Joules). Hint: use de Broglie's formula,
p = h/λ
to find the wavelength of the electrons.

We first need to calculate the wavelength of electrons with this energy. Assuming all this energy is kinetic we have
T = = 0.5×1.6×10-19
Joules. Thus
p = = 3.82×10-25
kgm/s. Then
λ = h/p = 6.626×10-32/3.82×10-25 = 1.74×10-9
meters. The distance between minima is the same as between any two maxima, so it will suffice to calculate the position of the first maximum. This is given by
y = = = = 4.34
millimeters.

Problem :
A Michelson interferometer can be used to calculate the wavelength of light by moving on of the mirrors and observing the number of fringes that move past a particular point. If a displacement of the mirror by
λ/2
causes each fringe to move to the position of an adjacent fringe, calculate the wavelength of light being used if 92 fringe pairs pass a point when the mirror is shifted
2.53×10-5
meters.

Since for each
λ/2
moved one fringe moves to the position of an adjacent one, we can deduce that the total distance moved
D
, divided by the number of displaced fringes
N
must be equal to
λ/2
. Thus:
D/N = λ/2
. Clearly, then
λ = 2D/N = = 5.50×10-7
meters, or 550 nanometers.